On Dec 30, 1:30*pm, David Nebenzahl wrote:
In a more-or-less recent thread up yonder (the one about LED lighting
that evolved into a discussion/argument about CFLs vs incandescents and
power factor, among other things), a technical term and concept (power
factor) was argued at length. I wonder how many folks actually were able
to follow those arguments.

Myself, I really didn't know just what this mysterious "power factor"
was. I did know that values lower than 1 were bad and caused power
distribution inefficiencies that resulted in real losses of energy and
money.

I now know what power factor is--sort of. The best explanation I ran
across on the web was this really simple one. Instead of taking the
mealy-mouthed Wikipedia approach of jumping in all cosines and formulae
phase angles and other fancy stuff and *then* explaining just what the
hell it *is*, this explanation is for the layperson:

* *Power factor in electricity is like efficiency. The best power factor
* *is 100%.

* *Consider a child on a swing. If you push them when they are going
* *backwards you will actually slow them down. In order to push with
* *maximum efficiency the motion of the swing and and your push must be
* *"in phase".

* *Similarly in electricity, voltage and current must be in phase for
* *optimum performance. Equipment such as motors, ballasts and variable
* *speed drives tends to move voltage and current out of phase with each
* *other.

[see athttp://www.carleton.ca/energy/powfac.htm]

Now that's the kind of explanation I like; simple and to the point. Of
course, the picky purist might object to the "best power factor is 100%"
thing (the best power factor is actually 1), but who cares? Now I
understand the concept.

So it turns out that PF is actually computed as the absolute cosine of
the phase angle, which also makes sense if one thinks about it. But I
still don't really have a handle on the meaning of this number. How low
does PF have to get before it's considered really bad? 0.8? 0.5? Don't
have much of a handle on that yet. (That's the problem with them
dimensionless numbers.)

I still don't know exactly how PF losses work in the real world, though
I can take an educated guess that they result mostly in heating in
transformers, transmission lines, etc.

--
I am a Canadian who was born and raised in The Netherlands. I live on
Planet Earth on a spot of land called Canada. We have noisy neighbours.

- harvested from Usenet

For a power factor of one, the voltage and current are in exact
agreement/phase for the peaks and nulls. If you apply a voltage to a
pure resistor, they are in phase and the power used is the product of
the voltage and current. If you apply a voltage to an inductor/motor/
transformer, the current lags behind the voltage somewhat and the
power into the device is not the product of the voltage times the
current, but the product of the voltage times the current times the
power factor. If you apply a voltage to a capacitor/condensor, the
current leads the voltage somewhat. Again, the actual power into the
device is the product of the voltage times the curent times the power
factor. If you have a very large inductance. the current will lag far
behind the voltage, and the actual power used in the device is small.
But, the power company still has to be able to provide the maximum
current to the device, but does not get much $$ since the power
atually consumed is much less. That is why the power companies do not
like inductive or capacitive loads. Compact fluorescent lights tend
to look like capacitive loads because they have a large inrush current
that leads the voltage and so altho the power companies will save some
power when these are adopted over a widespeard area, they will still
have to be able to provide the peak current that these lamps draw.
Fortunately, that peak curent is still well below what a resistive
incandescent light that puts out the same lumens uses.

Hope this helps.

hr(bob) wrote:

For a power factor of one, the voltage and current are in exact
agreement/phase for the peaks and nulls. If you apply a voltage to a
pure resistor, they are in phase and the power used is the product of
the voltage and current. If you apply a voltage to an inductor/motor/
transformer, the current lags behind the voltage somewhat and the
power into the device is not the product of the voltage times the
current, but the product of the voltage times the current times the
power factor. If you apply a voltage to a capacitor/condensor, the
current leads the voltage somewhat. Again, the actual power into the
device is the product of the voltage times the curent times the power
factor. If you have a very large inductance. the current will lag far
behind the voltage, and the actual power used in the device is small.
But, the power company still has to be able to provide the maximum
current to the device, but does not get much $$ since the power
atually consumed is much less. That is why the power companies do not
like inductive or capacitive loads. Compact fluorescent lights tend
to look like capacitive loads because they have a large inrush current
that leads the voltage and so altho the power companies will save some
power when these are adopted over a widespeard area, they will still
have to be able to provide the peak current that these lamps draw.
Fortunately, that peak curent is still well below what a resistive
incandescent light that puts out the same lumens uses.

Hope this helps.

If you study how a mechanical killowatt-hour meter works, you'll find that
it depends on the voltage and current being in phase. Usually phase-shifting
loads (transformers, motors, CHLs) are negligible (their resistive loads are
much greater than their reactive loads) and their effect is ignored. Put a
big enough reactive load (that shifts the phase between voltage and current
90°) on the circuit and the old-fashioned, mechanical, killowatt-hour meter
will do what?

It will stop. Or run backwards.

Specifically:

"Some combinations of capacitive and inductive load can interact with the
coils and mass of a rotor and cause reduced or reverse motion. All of these
effects can be detected by the electric company, and many modern meters can
detect or compensate for them.[nudge-nudge, wink-wink]" ["Detection" is
usually accomplished by looking at the WTF? box attached to the distribution
panel.]

hr(bob) wrote:

For a power factor of one, the voltage and current are in exact
agreement/phase for the peaks and nulls. If you apply a voltage to a
pure resistor, they are in phase and the power used is the product of
the voltage and current. If you apply a voltage to an inductor/motor/
transformer, the current lags behind the voltage somewhat and the
power into the device is not the product of the voltage times the
current, but the product of the voltage times the current times the
power factor. If you apply a voltage to a capacitor/condensor, the
current leads the voltage somewhat. Again, the actual power into the
device is the product of the voltage times the curent times the power
factor. If you have a very large inductance. the current will lag far
behind the voltage, and the actual power used in the device is small.
But, the power company still has to be able to provide the maximum
current to the device, but does not get much $$ since the power
atually consumed is much less. That is why the power companies do not
like inductive or capacitive loads. Compact fluorescent lights tend
to look like capacitive loads because they have a large inrush current
that leads the voltage and so altho the power companies will save some
power when these are adopted over a widespeard area, they will still
have to be able to provide the peak current that these lamps draw.
Fortunately, that peak curent is still well below what a resistive
incandescent light that puts out the same lumens uses.


I agree with other comments - particularly hr[bob], Wayne, George and
trader.

Accuracy of watt-hour meters comes up at alt.engineering.electrical.
Both mechanical and electronic meters accurately measure the real power
(watts). There is no question.
(If anyone is interested in how mechanical watt-hour meters work
details are in a thread on alt.engineer.electrical starting 11-21-09 and
titled "Balancing the Breaker Box". Much of the thread is forgettable,
but look at posts by and daestrom , particularly
starting 11-30. This is only readable by techies.)

Adding to hr[bob]'s post - the current through an inductor (as in a
motor) stores energy in a magnetic field. As the current is increasing,
additional current is required to store the energy. As the current is
decreasing the energy comes out as current that flows back into the
grid. This shifts the peaks of the current flow away from the peaks of
the voltage. For a motor, the overall current is increased over the
current that does mechanical work.

In a capacitor, energy is stored in an electric field. As with an
inductance, the energy flows into the field and back out to the grid,
shifting the voltage and current peaks.

Watt-hour meters ignore reactive currents (inductance and capacitance)
and measure only real power (watts). (They record energy - watt-hours.)

Energy storage in capacitors counter balances energy storage in
inductances and can return power factor toward 1. Utilities have racks
of PF correction capacitors to raise the PF toward 1, which reduces
currents, which reduces utility resistance losses.

Utilities can meter several things. One is real power (energy), which is
what watt-hour meters measure. For commercial and particularly
industrial facilities they may also measure "demand" and "VARs".

"Demand", which is in Daring's posts, is the peak power consumed (max of
watt-hours measured over a small time period). High peak power means the
utility has to have capacity to supply that demand. If usage can be
shifted, like cycling air conditioning when a facility is using a lot of
power in other areas, the capacity the utility has to provide is reduced
and the facility is rewarded by lower "demand" charges. IMHO motor
starts are too short to change demand much. On a mechanical meter demand
is a dial around the top of the face of the watt-hour meter.

"VARs" are volt-amps - reactive. This is reactive power, and peaks when
voltage and current are at 90 degrees. If V and A are in phase
(resistive load) the VARs are zero. It is measured with a separate
watt-hour meter (could be combined in an electronic meter - don't know
if they do). This reactive power is not "used" - it flows from the grid
and back to the grid 120 times per second. But it increases the
resistance losses for the utility, and the utility can charge a big
penalty for VAR 'use'. That provides an incentive for facilities to
reduce the VARs using PF correction capacitors which, as someone said,
have to be switched in and out depending on motor load. PF correction at
motors reduces the resistance losses within the facility.

Since residential users pay no VAR penalty "black boxes" to increase PF
are a scam. Residential users pay only for power that is used.

Equipment that includes a DC supply distorts the current, with high
current near the peak voltage and low or zero current the rest of the
cycle. This includes switch mode power supplies (computers, CFLs) and
variable speed drives. The V & A peaks are not so much shifted as the
current wave form is non-sinusoidal. I think this is referred to
displacement power factor. (Power factor is defined as power divided by
volts times amps.) I believe the European Union has limits on this.

--
bud--

In article ,
hr(bob) wrote:
....

- harvested from Usenet

For a power factor of one, the voltage and current are in exact
agreement/phase for the peaks and nulls. If you apply a voltage to a
pure resistor, they are in phase and the power used is the product of
the voltage and current. If you apply a voltage to an inductor/motor/
transformer, the current lags behind the voltage somewhat and the
power into the device is not the product of the voltage times the
current, but the product of the voltage times the current times the
power factor. If you apply a voltage to a capacitor/condensor, the
current leads the voltage somewhat. Again, the actual power into the
device is the product of the voltage times the curent times the power
factor. If you have a very large inductance. the current will lag far
behind the voltage, and the actual power used in the device is small.
But, the power company still has to be able to provide the maximum
current to the device, but does not get much $$ since the power
atually consumed is much less. That is why the power companies do not
like inductive or capacitive loads. Compact fluorescent lights tend
to look like capacitive loads because they have a large inrush current
that leads the voltage and so altho the power companies will save some
power when these are adopted over a widespeard area, they will still
have to be able to provide the peak current that these lamps draw.
Fortunately, that peak curent is still well below what a resistive
incandescent light that puts out the same lumens uses.

Hope this helps.

A not only much delayed but probably stupid question:

If you phase-shifted volt-amp pair, and multiplied them together
at each time-instant, and addeded them up (ie integrated the
product dt)?, would that give the same result as multiplying
the peak voltage and peak current by the pf?

Thanks,

David

(I gotta go read that wikipedia article!)

On 2010-01-30, David Combs wrote:

If you phase-shifted volt-amp pair,

I assume you mean voltage and current waveforms that may be out of
phase.

and multiplied them together at each time-instant, and addeded them
up (ie integrated the product dt)?, would that give the same result
as multiplying the peak voltage and peak current by the pf?

You have the right idea but the details are a little off. In the
numerator, you need to integrate over 1 second. And in the
denominator, you need to use the root-mean-square values of voltage
and current (which are the commonly referenced values anyway). Then
you'll have power factor. Otherwise, you'd be off by a constant
factor, i.e. you won't get a value of 1 when the voltage and current
are actually in phase.

Cheers, Wayne

On Jan 30, 3:19*pm, Wayne Whitney wrote:
On 2010-01-30, David Combs wrote:

If you phase-shifted volt-amp pair,

I assume you mean voltage and current waveforms that may be out of
phase.

and multiplied them together at each time-instant, and addeded them
up (ie integrated the product dt)?, would that give the same result
as multiplying the peak voltage and peak current by the pf?

You have the right idea but the details are a little off. *In the
numerator, you need to integrate over 1 second. *And in the
denominator, you need to use the root-mean-square values of voltage
and current (which are the commonly referenced values anyway). *Then
you'll have power factor. *Otherwise, you'd be off by a constant
factor, i.e. you won't get a value of 1 when the voltage and current
are actually in phase.

Cheers, Wayne

Actually he's right as far as using the instantaneous voltage and
current and doing the integration. That gives the true power. But
as you say, the comparison value calculated the normal value would be
to use the RMS voltage and current and power factor, not peak values.

On 2010-01-30, wrote:

Actually he's right as far as using the instantaneous voltage and
current and doing the integration. That gives the true power.

I think we are in agreement, I guess I just wasn't clear. If you
integrate V*A over a time interval, you get energy. So to get true
power, you need to divide by the length of the time interval. Which
is why I suggested integrating over an interval of 1 second.

Cheers, Wayne

On Jan 30, 7:11*pm, Wayne Whitney wrote:
On 2010-01-30, wrote:

Actually he's right as far as using the instantaneous voltage and
current and doing the integration. *That gives the true power. *

I think we are in agreement, I guess I just wasn't clear. *If you
integrate V*A over a time interval, you get energy. *So to get true
power, you need to divide by the length of the time interval. *Which
is why I suggested integrating over an interval of 1 second.

Cheers, Wayne


Yes, I agree. We're thinking the same thing, but I was a little off
there too After doing the integral, you do have to divide by the
time interval, whether it's one cycle or one sec, etc.